 # Minimum value of f(x)=(x^p)/(p)+(x^(-q))/(q) subjected to 1/p+1/q=1 and p>1 Although I have solved it using derivative. But did not understand how can i solve without derivative, Help Required, Thanks maredilunavy 2022-09-14 Answered
minimum value of $f\left(x\right)$ without the use of derivative
Although I have solved it using derivative. But did not understand how can i solve
without derivative, Help Required, Thanks
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Use Young's inequality.
$f\left(x\right)=\frac{{x}^{p}}{p}+\frac{{x}^{-q}}{q}\ge x\cdot \frac{1}{x}=1.$
Since $f\left(1\right)=1$, the minimum is $1$
###### Not exactly what you’re looking for? Krish Crosby
By AM-GM:
$\frac{{x}^{p}}{p}+\frac{{x}^{-q}}{q}\ge \frac{1}{p}+\frac{1}{q}+x+\frac{1}{x}-2\ge \frac{1}{p}+\frac{1}{q}=1$
Since ${x}^{p}-px+p-1\ge 0$ and ${x}^{-q}-\frac{q}{x}+q-1\ge 0$